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If $a_1=2,a_{n+1}=\sqrt{a_n+8}-\sqrt{a_n+3}$. Prove that $n\leq a_1+a_2+...+a_n \leq n+1$ for every $n\ge1$ and $\lim a_n=1$.


I have showed that by squaring and inequality techniques:

  1. $a_i<\sqrt{3}$ for every $i>1$.
  2. If $a_i>1$ then $a_{i+1}<1$ for every $i\ge 1$

I think that $\sqrt{3}$ can be improved, but I am not sure if it's useful.

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We have

$ a_{n+1}-1 = (a_n-1) \left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) $

and

$(a_{n+1}-1)+(a_n-1) = (a_n-1) \left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)$,

a) $ \left| \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+a_n}} + \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+0}} + \cfrac{1}{2+\sqrt{3+0}} \right| = 5-(\sqrt 3 + \sqrt8) < 1 $

This way we have $|a_{n+1}-1| < q |a_n-1| < q^n |a_1-1|$ by induction. This way, $|a_{n+1}-1|$ tends to 0.

b) We have to show some results:

  • $ \left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) < 0$, therefore $a_{n+1}-1$ and $a_n-1$ have opposite signals. But $a_1-1=1>0$, then the even guys are negative, the odd ones are positive: $a_{2k} < 1 < a_{2k+1}$.

  • $ \left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) > 0$, therefore $a_{n+1}+a_n-2$ and $a_n-1$ have the same signals: $a_{2k}+a_{2k+1} < 2 < a_{2k+1}+a_{2k+2}$

Now we have two cases to consider:

  • $ (a_1+a_2)+\ldots+(a_{2k+1}+a_{2k+2}) > 2+\ldots+2=2k+2$ and $ a_1+(a_2+a_3)+\ldots+(a_{2k}+a_{2k+1})+a_{2k+2} < 2+\ldots+2+1=2k+3$

  • $ (a_1+a_2)+\ldots+(a_{2k+1}) > 2+\ldots+1=2k+1$ and $ a_1+(a_2+a_3)+\ldots+(a_{2k}+a_{2k+1}) < 2+\ldots+2+2=2k+2$

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